Problem 29 - College Trigonometry

Question

We know that \(|x+y| \leq|x|+|y|\) for all real numbers \(x\) and \(y\) by the Triangle Inequality established in Exercise 36 in Section 2.2. We can now establish a Triangle Inequality for vectors. In this exercise, we prove that \(\|\vec{u}+\vec{v}\| \leq\|\vec{u}\|+\|\vec{v}\|\) for all pairs of vectors \(\vec{u}\) and \(\vec{v}\). (a) (Step 1) Show that \(\|\vec{u}+\vec{v}\|^{2}=\|\vec{u}\|^{2}+2 \vec{u} \cdot \vec{v}+\|\vec{v}\|^{2}\). 6 (b) (Step 2) Show that \(|\vec{u} \cdot \vec{v}| \leq\|\vec{u}\|\|\vec{v}\| .\) This is the celebrated Cauchy-Schwarz Inequality. (Hint: To show this inequality, start with the fact that \(|\vec{u} \cdot \vec{v}|=|\|\vec{u}\|\|\vec{v}\| \cos (\theta)|\) and use the fact that \(|\cos (\theta)| \leq 1\) for all \(\theta\).) (c) (Step 3) Show that \(\|\vec{u}+\vec{v}\|^{2}=\|\vec{u}\|^{2}+2 \vec{u} \cdot \vec{v}+\|\vec{v}\|^{2} \leq\|\vec{u}\|^{2}+2|\vec{u} \cdot \vec{v}|+\|\vec{v}\|^{2} \leq\|\vec{u}\|^{2}+\) \(2\|\vec{u}\|\|\vec{v}\|+\|\vec{v}\|^{2}=(\|\vec{u}\|+\|\vec{v}\|)^{2}\) (d) (Step 4) Use Step 3 to show that \(\|\vec{u}+\vec{v}\| \leq\|\vec{u}\|+\|\vec{v}\|\) for all pairs of vectors \(\vec{u}\) and \(\vec{v}\). (e) As an added bonus, we can now show that the Triangle Inequality \(|z+w| \leq|z|+|w|\) holds for all complex numbers \(z\) and \(w\) as well. Identify the complex number \(z=a+b i\) with the vector \(u=\langle a, b\rangle\) and identify the complex number \(w=c+d i\) with the vector \(v=\langle c, d\rangle\) and just follow your nose!

Step-by-Step Solution

Answer

The triangle inequality \(\|\vec{u}+\vec{v}\| \leq \|\vec{u}\|+\|\vec{v}\|\) holds for vectors and complex numbers.

Step 1: Understanding Vector Norms and Inner Products

The norm of a vector \(\vec{u}\) is \(\|\vec{u}\|=\sqrt{\vec{u}\cdot\vec{u}}\). The expression \(\|\vec{u}+\vec{v}\|^{2}\) can be expanded using the definition of the dot product: \(\|\vec{u}+\vec{v}\|^{2} = (\vec{u}+\vec{v})\cdot(\vec{u}+\vec{v}) = \vec{u}\cdot\vec{u} + 2 \vec{u}\cdot\vec{v} + \vec{v}\cdot\vec{v}\). Thus, \(\|\vec{u}+\vec{v}\|^{2} = \|\vec{u}\|^{2} + 2\vec{u}\cdot\vec{v} + \|\vec{v}\|^{2}\).

Step 2: Applying the Cauchy-Schwarz Inequality

To show \(|\vec{u} \cdot \vec{v}| \leq \|\vec{u}\|\|\vec{v}\|\), consider that \(\vec{u} \cdot \vec{v} = \|\vec{u}\| \|\vec{v}\| \cos(\theta)\), where \(\theta\) is the angle between the vectors. Since \(|\cos(\theta)| \leq 1\), this gives us \(|\vec{u} \cdot \vec{v}| = |\|\vec{u}\| \|\vec{v}\| \cos(\theta)| \leq \|\vec{u}\| \|\vec{v}\|\). This inequality is known as the Cauchy-Schwarz Inequality.

Step 3: Establishing the Inequality for Squared Norms

Using the result from Step 1, we have \(\|\vec{u}+\vec{v}\|^{2} = \|\vec{u}\|^{2} + 2 \vec{u}\cdot\vec{v} + \|\vec{v}\|^{2}\). Applying the Cauchy-Schwarz Inequality gives \(|\vec{u}\cdot\vec{v}| \leq \|\vec{u}\|\|\vec{v}\|\), so \(\|\vec{u}+\vec{v}\|^{2} \leq \|\vec{u}\|^{2} + 2\|\vec{u}\|\|\vec{v}\| + \|\vec{v}\|^{2}\). This is equal to \((\|\vec{u}\|+\|\vec{v}\|)^{2}\).

Step 4: Concluding with the Triangle Inequality

Since \(\|\vec{u}+\vec{v}\|^{2} \leq (\|\vec{u}\|+\|\vec{v}\|)^{2}\), taking the square root of both sides yields \(\|\vec{u}+\vec{v}\| \leq \|\vec{u}\| + \|\vec{v}\|\). This confirms the Triangle Inequality for vectors.

Step 5: Applying to Complex Numbers

Given complex numbers \( z = a + bi \) and \( w = c + di \), we can represent these as vectors \( \vec{u} = \langle a, b \rangle \) and \( \vec{v} = \langle c, d \rangle \). By applying the triangle inequality for vectors, \(|z+w| = \|\vec{u}+\vec{v}\| \leq \|\vec{u}\| + \|\vec{v}\| = |z| + |w|\), thus confirming the triangle inequality for complex numbers.

Key Concepts

Cauchy-Schwarz Inequality

The Cauchy-Schwarz Inequality is a fundamental concept in vector mathematics. It states that for any vectors \( \vec{u} \) and \( \vec{v} \), the absolute value of their dot product is less than or equal to the product of their magnitudes. Mathematically, this is expressed as:

• \( |\vec{u} \cdot \vec{v}| \leq \|\vec{u}\| \|\vec{v}\| \)

This inequality is crucial because it provides an upper bound for the dot product. To understand it, consider that the dot product \( \vec{u} \cdot \vec{v} \) represents the product of the lengths of the vectors and the cosine of the angle \( \theta \) between them:

• \( \vec{u} \cdot \vec{v} = \|\vec{u}\| \|\vec{v}\| \cos(\theta) \)

Since \( |\cos(\theta)| \leq 1 \), the inequality naturally follows. This concept is widely used in proving other mathematical inequalities and properties in vector spaces.

Complex Numbers

Complex numbers extend the concept of the one-dimensional number line by including imaginary numbers, which are multiples of \( i \), the imaginary unit defined as \( i^2 = -1 \). A complex number \( z \) takes the form \( z = a + bi \), where \( a \) is the real part and \( b \) is the imaginary part.

In terms of geometric interpretation, you can represent them as vectors in a 2-D plane:

• \( z = a + bi \) can be identified with the vector \( \langle a, b \rangle \)

The norm, or modulus, of a complex number is given by \( |z| = \sqrt{a^2 + b^2} \).

The Triangle Inequality for complex numbers states that for any complex numbers \( z \) and \( w \), the inequality \( |z+w| \leq |z| + |w| \) holds. By relating complex numbers to vectors, this concept was demonstrated using the Triangle Inequality for vectors.

Vector Norms

A vector norm is a measure of the length or magnitude of a vector. For a vector \( \vec{u} \), its norm is denoted by \( \|\vec{u}\| \) and is calculated as the square root of the dot product of the vector with itself:

• \( \|\vec{u}\| = \sqrt{\vec{u} \cdot \vec{u}} \)

This concept is fundamental in linear algebra and helps in comparing the sizes of vectors. The norm is always non-negative and only equals zero when the vector itself is zero.

Vector norms are essential when proving the Triangle Inequality for vectors, which extends the principle of the classical Triangle Inequality from real numbers to vectors, providing a way to understand how the lengths of vectors relate when they are added together.

Dot Product

The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. For vectors \( \vec{u} = \langle u_1, u_2, \ldots, u_n \rangle \) and \( \vec{v} = \langle v_1, v_2, \ldots, v_n \rangle \), the dot product is given by:

• \( \vec{u} \cdot \vec{v} = u_1v_1 + u_2v_2 + \ldots + u_nv_n \)

The dot product has several key properties:

• It is commutative: \( \vec{u} \cdot \vec{v} = \vec{v} \cdot \vec{u} \)
• It distributes over vector addition, meaning it respects the operations of addition and multiplication, just like regular algebra.

An important use of the dot product is determining the angle \( \theta \) between two vectors, through the formula \( \cos(\theta) = \frac{\vec{u} \cdot \vec{v}}{\|\vec{u}\| \|\vec{v}\|} \). When the dot product is zero, it indicates that the vectors are orthogonal, meaning they are at a right angle to each other.